IODS course project
siqizhao
Introduction to Open Data Science - Course Project
About the project
Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.
# This is a so-called "R chunk" where you can write R code.
date()## [1] "Mon Nov 23 10:09:29 2020"The text continues here.
Learning R is an interesting process even though there are a lot of difficulties which can not be sloved sometimes. I hope I could learn more and more in this course.I found this course in weboodi.
Regression and model validation
This week I learned how to use R to wrangling data and make the linear regression analysis and evaluate the validation of model.
- Describe your work and results clearly.
- Assume the reader has an introductory course level understanding of writing and reading R code as well as statistical methods.
- Assume the reader has no previous knowledge of your data or the more advanced methods you are using.
Here we go again…
In the exercise 2, the Code is as following:
This data files includes 166 rows (observations) and 7 columns(variables), which means there are 7 kinds of index for 166 students.The variables includes gender, age, attuitude, deep, dtra, surf, and points.
students2014 <- read.table('http://s3.amazonaws.com/assets.datacamp.com/production/course_2218/datasets/learning2014.txt', sep = ',', header = TRUE)
dim(students2014)## [1] 166 7str(students2014)## 'data.frame': 166 obs. of 7 variables:
## $ gender : chr "F" "M" "F" "M" ...
## $ age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ attitude: num 3.7 3.1 2.5 3.5 3.7 3.8 3.5 2.9 3.8 2.1 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ points : int 25 12 24 10 22 21 21 31 24 26 ...library(ggplot2)## Warning: package 'ggplot2' was built under R version 4.0.2p1 <- ggplot(students2014, aes(x=attitude, y=points, col=gender))
p2 <- p1 + geom_point()
p2
p3 <- p2+geom_smooth(method = 'lm')
p4 <- p3+ggtitle('attitude of students versus exam points')
p4## `geom_smooth()` using formula 'y ~ x'
pairs(students2014[-1])
library(GGally)## Warning: package 'GGally' was built under R version 4.0.3## Registered S3 method overwritten by 'GGally':
## method from
## +.gg ggplot2library(ggplot2)
p <- ggpairs(students2014, mapping=aes(col=gender, alpha=0.3),lowe=list(combo=wrap('facethist', bins=20)))
p
There is an extremely significant positive relationship between attitude of students and points.For male,age is negative related with points.
my_model1 <- lm (points ~ attitude + stra + surf, data = students2014)
summary(my_model1)##
## Call:
## lm(formula = points ~ attitude + stra + surf, data = students2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.0171 3.6837 2.991 0.00322 **
## attitude 3.3952 0.5741 5.913 1.93e-08 ***
## stra 0.8531 0.5416 1.575 0.11716
## surf -0.5861 0.8014 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08my_model2 <- lm (points ~ attitude + stra, data = students2014)
summary(my_model2)##
## Call:
## lm(formula = points ~ attitude + stra, data = students2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.6436 -3.3113 0.5575 3.7928 10.9295
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 8.9729 2.3959 3.745 0.00025 ***
## attitude 3.4658 0.5652 6.132 6.31e-09 ***
## stra 0.9137 0.5345 1.709 0.08927 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.289 on 163 degrees of freedom
## Multiple R-squared: 0.2048, Adjusted R-squared: 0.1951
## F-statistic: 20.99 on 2 and 163 DF, p-value: 7.734e-09In the first regression model, points as the target varable together with attitude, stra and surf as explantory variables, the model can be used because the P value is less than 0.001(extremely significant) even though there is no significantly relationships between stra, surf and points.
Model2 can explain the 20.74% changes of points.
par(mfrow=c(2,2))
plot(my_model2, which=c(1,2,5))
The data is mostly Gaussian distribution, according to the linear qq plots. The model is sensible based on these three plots.
Logistic Regression and cross validation
alc <- read.table('http://s3.amazonaws.com/assets.datacamp.com/production/course_2218/datasets/alc.txt', sep=',', header = TRUE)In this data set, there are 35 variables and 382 observations in total.The variables are: school,sex,age,address,famsize,Pstatus,Medu,Fedu,Mjob,Fjob reason,nursery,internet,guardian,traveltime studytime,failures,schoolsup,famsup,paid activities,higher,romantic,famrel,freetime goout,Dalc,Walc,health,absences,G1,G2,G3,alc_use,high_use.
I chose failures, absences,traveltime and activities as explainary variables to explore the relationsips between them and the target variable-high_use.The results are as below(three of them have significatly positive relationship with high_use): high_use=0.4176failures-1.0157 high_use=0.0683absences-1.2640 high_use=0.4290*traveltime-1.5146
So, the high_use valus will increase with the increase of failures, absences and traveltime. However, there is no significant relationship between activities and high_use.
library(ggplot2)
library(tidyr)## Warning: package 'tidyr' was built under R version 4.0.2library(dplyr)## Warning: package 'dplyr' was built under R version 4.0.2##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, uniongather(alc) %>% glimpse## Rows: 13,370
## Columns: 2
## $ key <chr> "school", "school", "school", "school", "school", "school", "...
## $ value <chr> "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", "GP", "...gather(alc) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
g1 <- ggplot(alc, aes(x = high_use, y = failures, col = sex))
g1 + geom_boxplot() + ylab("failures")
g2 <- ggplot(alc, aes(x = high_use, y = absences, col = sex))
g2 + geom_boxplot() + ggtitle("Student absences by alcohol consumption and sex")
If we look at the combined bar chart, we can find the numbers of students who attend extra-curricular activities and the students who don’t are almost close, which can tell us the non-significat relationship between activities and high_use is sensible.Let’s use the same principle to explain the other relationships, and the results should be consistent with the original hypothese.
Based on the results of Logistic regression analysis between failures&absences&traveltime&activities and high_use, the fitted model is as below:
(Intercept) failures absences traveltime activitiesyes -1.86181552 0.36618544 0.07004812 0.42610620 -0.32180313
OR 2.5 % 97.5 % (Intercept) 0.1553903 0.0828279 0.2841262 failures 1.4422227 1.0755782 1.9361180 absences 1.0725598 1.0374448 1.1130908 traveltime 1.5312834 1.1164457 2.1057338 activitiesyes 0.7248409 0.4542950 1.1528149
high_use= 0.36619 * failures+0.07005 * absences + 0.42611 * travelime -1.86182
There is no significat relationship between activities and high_use, which is same with the previous hypothese, and should be removed fromt the whole model.
The 2x2 cross tabulation of predictions is as below:
prediction
high_use FALSE TRUE
FALSE 261 9
TRUE 87 25
m5 <- glm(high_use ~ failures+absences+traveltime, data = alc, family = "binomial")
probabilities <- predict(m5, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
select(alc, failures, absences, traveltime, high_use, probability, prediction) %>% tail(10)## failures absences traveltime high_use probability prediction
## 373 1 0 2 FALSE 0.3105560 FALSE
## 374 1 14 2 TRUE 0.5424959 TRUE
## 375 0 2 2 FALSE 0.2616419 FALSE
## 376 0 7 3 FALSE 0.4322838 FALSE
## 377 1 0 1 FALSE 0.2285089 FALSE
## 378 0 0 1 FALSE 0.1686879 FALSE
## 379 1 0 2 FALSE 0.3105560 FALSE
## 380 1 0 2 FALSE 0.3105560 FALSE
## 381 0 3 2 TRUE 0.2752164 FALSE
## 382 0 0 3 TRUE 0.3194072 FALSEtable(high_use = alc$high_use, prediction = alc$prediction)## prediction
## high_use FALSE TRUE
## FALSE 260 10
## TRUE 88 24g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))
g + geom_point()
table(high_use = alc$high_use, prediction = alc$prediction) %>% prop.table %>% addmargins## prediction
## high_use FALSE TRUE Sum
## FALSE 0.68062827 0.02617801 0.70680628
## TRUE 0.23036649 0.06282723 0.29319372
## Sum 0.91099476 0.08900524 1.00000000According to the 10-folds cross-validation on your mode, I got a smaller error (0.2591623) than Datacamp, which means that my model has better test set performance.
I made two different fited model with more predictors;model6 has 5 predictors which are failures,absences,traveltime,higher as well as sex; while model 7 has 4 explainary factors.
The 10 folds cross validation results are as below:
m6-> 0.2408377 m7-> 0.2382199
Compared with m5 (0.2591623), the errors of m7 is the smallest one which has one more predictor than m5.The fitted model is:
(Intercept) failures absences traveltime sexM -2.51975566 0.36718070 0.07462037 0.39558320 0.96465837
high_use= 0.36718070 * failures + 0.07462037 * absences + 0.39558320 * traveltime + 0.96465837 * sexM -2.51975566
Clustering and classification
The boston data frame has 506 rows and 14 columns, which means that there are 14 variables. crim is per capita crime rate by town; zn is proportion of residential land zoned for lots over 25,000 sq.ft.; indus is the proportion of non-retail business acres per town; chas is Charles River dummy variable (= 1 if tract bounds river; 0 otherwise); nox is nitrogen oxides concentration (parts per 10 million); rm is average number of rooms per dwelling; age is proportion of owner-occupied units built prior to 1940; dis is weighted mean of distances to five Boston employment centres;rad is index of accessibility to radial highways; tax is full-value property-tax rate per $10,000; ptratio is pupil-teacher ratio by town.black is 1000(Bk - 0.63)^2 where Bk is the proportion of blacks by town; lstat means lower status of the population (percent); medv means median value of owner-occupied homes in $1000s.
According to the correlation matrix, there is an obverse positive relationship between rad and tax, indus and nox, while there is an negative relationship between dis and nox, indus as well as dis. Also, seems like there are no obverse correlations between chas and all the others.
# access the MASS package
library(MASS)## Warning: package 'MASS' was built under R version 4.0.3##
## Attaching package: 'MASS'## The following object is masked from 'package:dplyr':
##
## select# load the data
data("Boston")
# center and standardize variables
boston_scaled <- scale(Boston)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865# class of the boston_scaled object
class(boston_scaled)## [1] "matrix" "array"# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
# summary of the scaled crime rate
summary(boston_scaled$crim)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## -0.419367 -0.410563 -0.390280 0.000000 0.007389 9.924110# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# remove original crim from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)After scaling the dataset, all the numbers become much smaller than oringinal ones and all the mean values are zero.
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# print the lda.fit object
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2500000 0.2450495 0.2475248 0.2574257
##
## Group means:
## zn indus chas nox rm age
## low 0.90265828 -0.9666313 -0.07742312 -0.8722582 0.43292068 -0.8954392
## med_low -0.06539846 -0.3197867 -0.03371693 -0.5860247 -0.16464855 -0.3090766
## med_high -0.38347752 0.1965965 0.20012296 0.3838094 0.07993772 0.4279782
## high -0.48724019 1.0149946 -0.08304540 1.0645106 -0.39752441 0.8167465
## dis rad tax ptratio black lstat
## low 0.9508873 -0.7010085 -0.7335666 -0.45188475 0.36630332 -0.76521528
## med_low 0.4282093 -0.5445257 -0.4700867 -0.02378498 0.31176268 -0.10969886
## med_high -0.4077193 -0.4352009 -0.3191413 -0.30510403 0.06007642 0.04763112
## high -0.8613193 1.6596029 1.5294129 0.80577843 -0.84622875 0.90620472
## medv
## low 0.52479558
## med_low -0.03530737
## med_high 0.13288897
## high -0.69692909
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.184478116 0.49641843 -0.99256208
## indus 0.003893237 -0.35071073 0.46663638
## chas -0.104095802 -0.01378560 0.09256502
## nox 0.248475006 -0.83668554 -1.39882157
## rm -0.150942077 -0.12351181 -0.22735696
## age 0.212316274 -0.36725249 0.10528519
## dis -0.229357720 -0.12558982 0.38227755
## rad 3.724733585 0.82076081 0.04787377
## tax 0.025834834 0.24831857 0.36339482
## ptratio 0.146741292 0.03487187 -0.26542221
## black -0.145511066 0.01113759 0.11684049
## lstat 0.133175486 -0.13366613 0.28675253
## medv 0.146798760 -0.26042455 -0.16308941
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9602 0.0300 0.0099# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 17 7 2 0
## med_low 7 11 9 0
## med_high 0 10 14 2
## high 0 0 1 22r = getOption("repos")
r["CRAN"] = "http://cran.us.r-project.org"
options(repos = r)
# access the MASS package
library(MASS)
data('Boston')
# center and standardize variables
boston_scaled <- scale(Boston)
# euclidean distance matrix
dist_eu <- dist(boston_scaled)
# look at the summary of the distances
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970# manhattan distance matrix
dist_man <- dist(boston_scaled, method = 'manhattan')
# look at the summary of the distances
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
## Bonus
# k-means clustering
km <-kmeans(boston_scaled, centers = 3)
# plot the Boston dataset with clusters
pairs(boston_scaled, col = km$cluster)
## Superbonus
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
install.packages("plotly")## Installing package into 'C:/rlibs/4.0.1'
## (as 'lib' is unspecified)## package 'plotly' successfully unpacked and MD5 sums checked
##
## The downloaded binary packages are in
## C:\Users\siqizhao\AppData\Local\Temp\RtmpCy92ha\downloaded_packagesplotly::plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color=train$crime)## Warning: `arrange_()` is deprecated as of dplyr 0.7.0.
## Please use `arrange()` instead.
## See vignette('programming') for more help
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_warnings()` to see where this warning was generated.